Answer:
Part A) [tex]h(4)-m(16)=-4[/tex]
Part B) The Y- intercepts are four units apart.
Part C) m(x) never exceeds h(x)
Step-by-step explanation:
Part A) What is the value of h(4)-m(16)
step 1
Find the value of h(4)
we have
[tex]h(x)=\frac{1}{2}(x-2)^2[/tex]
For x=4
substitute in the quadratic equation
[tex]h(4)=\frac{1}{2}(4-2)^2[/tex]
[tex]h(4)=2[/tex]
step 2
Find the equation of m(x)
Find the slope m
The formula to calculate the slope between two points is equal to
[tex]m=\frac{y2-y1}{x2-x1}[/tex]
take two points from the data
(8,2) and (10,3)
substitute in the formula
[tex]m=\frac{3-2}{10-8}[/tex]
[tex]m=\frac{1}{2}[/tex]
Find the equation in point slope form
[tex]y-y1=m(x-x1)[/tex]
we have
[tex]m=\frac{1}{2}[/tex]
[tex]point\ (8,2)[/tex]
substitute
[tex]y-2=\frac{1}{2}(x-8)[/tex]
Convert to slope intercept form
[tex]y=mx+b[/tex]
Isolate the variable y
[tex]y-2=\frac{1}{2}x-4[/tex]
[tex]y=\frac{1}{2}x-2[/tex]
Convert to function notation
[tex]m(x)=\frac{1}{2}x-2[/tex]
Find the value of m(16)
For x=16
substitute the value of x in the function m(x)
[tex]m(16)=\frac{1}{2}(16)-2[/tex]
[tex]m(16)=6[/tex]
Find h(4)-m(16)
substitute the values
[tex]2-6=-4[/tex]
therefore
[tex]h(4)-m(16)=-4[/tex]
Part B) If both of these functions are graphed on the same coordinate plane, how far apart are the y-intercepts?
we know that
The y-intercept is the value of y when the value of x is equal to zero
Find the y-intercept of h(x)
For x=0
[tex]h(x)=\frac{1}{2}(0-2)^2=2[/tex]
The y-intercept of h(x) is the point (0,2)
Find the y-intercept of m(x)
For x=0
m(x)=\frac{1}{2}(0)-2=-2
The y-intercept of m(x) is the point (0,-2)
To find out how far apart are the y-intercepts, determine the difference between the y-intercepts
[tex]2-(-2)=4\ units[/tex]
see the attached figure to better understand the problem
Part C) What values of x does m(x) excess h(x)?
we know that
Looking at the attached figure
The graphs of f(x) and m(x) not intersect
The graph of f(x) is above the graph of m(x)
The value of f(x) will always be greater than the value of m(x)
therefore
m(x) never exceeds h(x)
